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The upper-left block is a planar rotation matrix. Its real characteristic polynomial is $(\lambda -\cos \theta {)}^2+{\sin }^2\theta$, which is strictly positive for real $\lambda$ because the angle excludes integer multiples of $\pi$. Hence any real eigenvector has $x=y=0$. The remaining equation is $5z=\lambda z$, givingThere are infinitely many nonzero solutions, spanning one independent eigenvector direction.

The characteristic polynomial is ${\lambda }^2-2\lambda \cos \theta +1$, with eigenvalues $\cos \theta \pm i\sin \theta$. A real diagonal form can have only real eigenvalues, so it requires $\sin \theta =0$. Conversely, when $\theta =k\pi$, the rotation matrix is already $(-1{)}^{k}I$. Thus

A three-dimensional rotation matrix is an orthogonal matrix with determinant $+1$. Here $\det A=b(a^2-d^2)$. For the branch $d=0$, the determinant condition gives $b=1$; for the branch $a=0$, it gives $b=-1$. HenceThe first is the identity rotation; the remaining three are half-turns. Their fixed axes are respectively the $y$ axis, the direction $(1,0,-1)$ and the direction $(1,0,1)$. The identity is included under the usual rotation matrix convention.

Among the two orthogonal matrices from part (d)(i), $d=-1$ gives $L=-I$, a rotation matrix through angle $\pi$. For $d=1$, the determinant is $-1$, so the matrix reverses orientation and is a reflection rather than a rotation:It is the reflection matrix across the $y$ axis.

The chain rule expresses the coordinate differential operators asThe coefficients are constant, so a second application givesandAdding cancels the mixed partial derivatives, and $a^2+b^2=1$ givesThe transformation has rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, so this identity expresses the rotational invariance of the Laplacian.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.

Let $J=A/a$, so $J^2=-I$. The even and odd powers in the matrix exponential giveHere the two scalar sums are the Taylor series for cosine and sine. ConsequentlyWith the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by $-a$; the parameter in the form printed in the paper is $\theta =a$.

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

Use the quarter-turn rotation matrixFor a general second-order tensor, direct multiplication givesRequiring $RT{R}^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish andFor an arbitrary rotation angle $\theta$, the transverse rotation is $Q_{\theta }=\cos \theta I+\sin \theta J$, where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_{\theta }$. Therefore $Q_{\theta }(\alpha I-\omega J)Q_{\theta }^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.

Hence invariance holds under every rotation about the axis, not just the quarter-turn. This is the general axisymmetric second-order tensor; a symmetric tensor would additionally require $\omega =0$, but that restriction was not assumed.

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

With the same right-handed active rotation convention as $R_z$, the $y$ direction is fixed, ${\mathbf{e}}_z$ rotates towards ${\mathbf{e}}_x$, and ${\mathbf{e}}_x$ towards $-{\mathbf{e}}_z$. ThereforeIts columns give the images of the three Cartesian unit vectors, which also makes the signs of this rotation matrix explicit.

A product of orthogonal matrices is orthogonal. In fact,Thus $\boxed{R_1\text{ is orthogonal for every }\theta ,\varphi .}$ Its determinant is also one, since each factor has determinant one, so it remains a proper rotation matrix.

Applying the rightmost rotation matrix first givesEquality with ${\mathbf{e}}_y$ requires $\cos \theta =0$, so $\theta =\pi /2$ in the specified range. Then $\sin \varphi =1$ and $\cos \varphi =0$ giveIn the reverse order, $R_z$ first fixes ${\mathbf{e}}_z$, so $R_2{\mathbf{e}}_z=R_y(\pi /2){\mathbf{e}}_z={\mathbf{e}}_x$. Therefore $\boxed{R_2\mathbf{v}\ne {\mathbf{v}}^{\prime }}$. This explicitly demonstrates that rotations about different axes do not generally commute.

For a Cartesian change of components $x_i^{\prime }=R_{ij}{x}_j$, a rank-$n$ tensor transforms according towith repeated indices summed. An isotropic tensor has unchanged components under every proper rotation matrix.

For rank two write $T^{\prime }=RT{R}^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turnsEach off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. ThereforeConversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the isotropic second-rank tensor classification without assuming that the original tensor was symmetric. Also $\lambda =T_{kk}/3$ is a scalar, since the trace is rotation-invariant.

The upper-left block of $U$ is the planar rotation matrix $R(\pi /2)$, and $U$ fixes the third coordinate. ThereforeThe diagonal entries of $V$ square to one, so $V^2=I$. Direct conjugation gives $VUV=U^{-1}$, hence $V{U}^{p}V=U^{-p}$ for every integer $p$. ConsequentlyWhen $p\mathrm{mod}4=0$, the last exponent is four, which is equivalent to zero because $U^4=I$.

The integration domain is invariant under every rotation matrix, and its volume element is unchanged. The substitution $\mathbf{y}=R\mathbf{x}$ therefore provesThus $I$ is an isotropic tensor. It is also completely symmetric in its four indices, because the coordinate factors commute. In the general isotropic form, exchanging indices $j,k$ equates $\alpha$ and $\beta$, and exchanging $j,l$ equates $\alpha$ and $\gamma$. Therefore

For a Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$ with $R$ an orthogonal matrix, a rank-$n$ tensor obeyswith repeated indices summed according to Einstein notation. A tensor is isotropic if $T_{i_1\cdots i_n}^{\prime }=T_{i_1\cdots i_n}$ for every proper rotation matrix $R$. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.

Under a proper rotation matrix $R$, the product of coordinates transforms as a rank-$n$ tensor. The unit ball is unchanged by the substitution $\mathbf{y}=R\mathbf{x}$, whose Jacobian determinant is one. HenceThus $S$ is an isotropic tensor. An isotropic second-rank tensor has the form $S_{ij}=A{\delta }_{ij}$. Taking the trace and using spherical coordinates,Therefore the isotropic second moment of a ball isAn isotropic third-rank tensor has the form $B{ϵ}_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf{x}\mapsto -\mathbf{x}$ makes its integrand odd on the ball. Thus

For a rotation matrix $R\in SO(3)$, the Cartesian tensor component transformation law isAn isotropic tensor is unchanged under every such rotation, so its components satisfy $T_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}$ for every $R\in SO(3)$. The repeated indices are summed by the Einstein summation convention.

Choose the following consistent sign convention. The Givens rotation $J(i,j,\theta )$ equals the identity matrix except for the block on coordinates $i,j$:Thus it changes only the $i$th and $j$th coordinates of a vector. Write $c=\cos \theta$, $s=\sin \theta$. Its two-coordinate block satisfiesAll remaining coordinates are unchanged, so $J^{T}J=I_n$. Hence every such Givens rotation is an orthogonal matrix. With this convention the two-coordinate rotation matrix rotates through angle $-\theta$.